Rivers, canals, drains and sewers flowing partly full are open channels — driven by gravity, with a free surface at atmospheric pressure. The workhorse design equation is Manning's equation, used worldwide for sizing channels and estimating discharge. This article explains it with worked examples.
Open Channel Flow Basics
Unlike pipe flow (driven by pressure), open channel flow is driven by the bed slope and has a free surface. Uniform flow occurs when depth and velocity are constant along the channel — the gravity component balances the boundary friction.
Manning's Equation
V = (1/n) · R^(2/3) · S^(1/2) Q = A · V
| Symbol | Meaning |
|---|---|
| V | Mean velocity (m/s) |
| n | Manning's roughness coefficient (dimensionless) |
| R = A/P | Hydraulic radius (m) |
| S | Bed slope (m/m) |
| A | Flow area; P = wetted perimeter |
Typical Manning's n Values
| Surface | Manning's n |
|---|---|
| Smooth concrete / cement plaster | 0.012 – 0.014 |
| Brickwork / ordinary concrete | 0.015 – 0.017 |
| Earthen channel, clean | 0.018 – 0.025 |
| Natural stream with weeds/stones | 0.030 – 0.050 |
Worked Example 1 — Discharge in a Rectangular Channel
A rectangular concrete channel is 3 m wide, flowing 1 m deep. Bed slope S = 0.001, n = 0.015. Find the velocity and discharge.
- Area A = b·y = 3 × 1 = 3 m²
- Wetted perimeter P = b + 2y = 3 + 2 = 5 m
- Hydraulic radius R = A/P = 3/5 = 0.6 m
- V = (1/0.015) × 0.6^(2/3) × 0.001^(1/2) = 66.67 × 0.7114 × 0.03162 = 1.50 m/s
- Q = A·V = 3 × 1.50 = 4.5 m³/s
Worked Example 2 — Slope Required for a Target Discharge
The same channel must carry 6 m³/s at 1 m depth. Find the bed slope required.
- Required V = Q/A = 6/3 = 2.0 m/s
- From Manning: S = (V·n / R^(2/3))² = (2.0 × 0.015 / 0.7114)² = (0.04218)² = 0.00178 (≈ 1 in 562)
Most Economical Rectangular Section
For a given area, the most efficient rectangular channel minimises the wetted perimeter. Setting dP/dy = 0 with A = b·y constant gives:
b = 2y and R = y/2
This carries the maximum discharge for the least lining — a key result for economical canal design. (For a trapezoidal channel, the most economical section is a half-hexagon.)
Worked Example 3 — Most Economical Section
Design the most economical rectangular concrete channel (n = 0.015) to carry 5 m³/s on a slope of 0.0009.
- Most economical: b = 2y, R = y/2, A = b·y = 2y²
- Q = A·V = 2y² × (1/n)(y/2)^(2/3) S^(1/2)
- 5 = 2y² × (1/0.015)(y/2)^(2/3)(0.0009)^(1/2) = 2y² × 66.67 × (y/2)^(2/3) × 0.03
- 5 = 4.0 y² (y/2)^(2/3) → solving, y ≈ 1.07 m, b = 2y ≈ 2.14 m
Applications
- Irrigation canal and field-channel design.
- Storm-water drains, road-side drains and culvert sizing.
- Sewer design (running partly full) and natural-stream flow estimation.
Common Mistakes
- Using the full perimeter (including the free surface) as wetted perimeter — the water surface is not a wetted boundary.
- Confusing hydraulic radius R with hydraulic depth or pipe radius.
- Inconsistent units — Manning's n above is for SI; US customary units carry a 1.49 factor.